Question

The squared pictured below has side lengths of 4 units. Questions are in the picture below-

Answer 1

A.

The length of the diagonal is given by the Pythagorean theorem therefore

`d=\sqrt[]{4^2+4^2}=\sqrt[]{16+16}=\sqrt[]{32}`

The length of the diagonal is ) units

B.

The area of the square is given by the next formula

`A=s^2`

where s is the side

s=4

`A=(4)^2=16units^2`

The area of the square is 16 units^2

C.

For the area of the triangle we will use the next formula

`A=(1)/(2)b* h`

where b is the base and h is the height

b=4 units

h=4units

`A=(1)/(2)(4)(4)=(1)/(2)(16)=8units^2`

The area of the triangle formed by a diagonal and two of the sides is 8 units^2

D.

For the area of this triangle, we will use the same formula that we use in C. but in this case

b=sqrt(32)/2

h=sqrt(32)/2

We substitute the values

`A=(1)/(2)(\frac{\sqrt[]{32}}{2})(\frac{\sqrt[]{32}}{2}))=4units^2`

The area of one of these triangles is 4 units^2

ANSWER

A. sqrt(32) units

B.16 units^2

C. 8 units^2

D. 4 units^2